Question

If the matrix A is of order 3x3 and the system of equations AX = B has a unique solution, what can be concluded about the determinant of A?

• A. The determinant of A is zero
• B. The determinant of A is non-zero
• C. The determinant of A must be 1 only
• D. The determinant of A cannot be negative

Hint:

For a system of linear equations \(AX=B\):

• If \(\det(A) \neq 0\), there is a unique solution.

• If \(\det(A) = 0\) and \((\text{adj } A)B \neq 0\), there is no solution (inconsistent system).

• If \(\det(A) = 0\) and \((\text{adj } A)B = 0\), there are infinitely many solutions (consistent system).
This summary is crucial for solving problems related to the nature of solutions of linear equations.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are dealing with a system of linear equations represented in matrix form as AX = B, where A is a 3x3 coefficient matrix. The question states that this system has a unique solution and asks about the property of the determinant of matrix A.

Step 2: Key Formula or Approach (Alternate Method):
Instead of using Cramer's rule, we can use the Invertible Matrix Theorem, which states that a system AX=B has a unique solution if and only if the coefficient matrix A is invertible.

Step 3: Detailed Explanation:
From the Invertible Matrix Theorem, the following statements are equivalent: 1. A is invertible. 2. det(A) ≠ 0. 3. The system AX=B has exactly one unique solution. If det(A) = 0, then A is singular and not invertible. In that case, AX=B has either no solution or infinitely many solutions. Since the problem states the system has a unique solution, A must be invertible. For A to be invertible, its determinant must be non-zero.

Step 4: Final Answer:
For the system of equations AX = B to have a unique solution, the determinant of the coefficient matrix A must be non-zero.